Michael's triangle

Geometry Level 4

In quadrilateral $ABCD,$ $AB = AD = 1,$ $BD = \sqrt{2},$ $BC = \dfrac{ \sqrt{2} }{ 2 },$ and $CD = \dfrac{ \sqrt{6} }{ 2}.$ The area of triangle $ABC$ can be expressed in the form $\dfrac{\sqrt{a} + b}{c}$ , where $a,$ $b,$ and $c$ are positive integers and $a$ is square-free. What is the value of $a+b+c$ ?

This problem is posed by Michael T.